Organizing National Elections in India to Elect the 543 Members of the Lok Sabha

Algorithmic Operations Research Vol.7 (2013) 55–70

Bodhibrata Nag a

aIndian Institute of Management Calcutta, Joka, Diamond Harbour Road, Kolkata, 700104, India Katta G. Murty b

bDepartment of IOE, University of Michigan, Ann Arbor, MI 48109-2117, USA

Abstract

There are 833 thousand polling stations in all of the 543 parliamentary constituencies spread over 35 states of India. On the day elections are being held in any one of these polling stations, a minimum of 4 Central Police Force(CPF) personnel must be deployed there, to maintain law and order and guarantee that voters can vote freely without being intimidated by anyone. As the number of CPF personnel available for this activity is limited, it is not possible to hold the Indian General elections on a single day over the whole country. So the set of 35 States of India is partitioned into a number of subsets, with elections in each subset of states being held on a single day. This partition is required to satisfy the constraints that the states in each subset are contiguous, and the subsets themselves must be contiguous. We present a method for organizing the Indian General Elections subject to these constraints, and minimizing the total number of election days required, and the total cost for the movement of CPF personnel involved. The method is based on the shortest Hamiltonian path problem, a tour segmentation problem deﬁned in the paper, and the bipartite minimum cost ﬂow problem.

Key words: OR in government; Scheduling; Graph partitioning; Hamiltonian path problem; tour segmentation problem; minimum cost ﬂow

1. Brief History of National Polls in India. of the Lok Sabha are directly elected by universal adult franchise by the electorate of all the 35 States through Three years after gaining independence from Britain, the “National Elections”, called General Elections in Bharat (India) became the Bharat Ganrajya (The Re- India. The term of office each Lok Sabha is five years public of India, or Indian Republic) by adopting the from the date of its first meeting, unless dissolved earlier Constitution of India in 1950. Now the Indian Repub- due to the ruling party losing a vote on a no-confidence lic comprises 28 States and 7 Union Territories, which motion in the Lok Sabha. These General Elections have we will refer to as the 35 States of India in the sequel. been held for the first time in the history of India in The Indian parliamentary form of governmentis federal 1951-52 after the adoption of the constitution of India; in structure with legislative powers distributed between and regularly after that as depicted in Table 1. In this the Parliament of India and State Legislatures. The Par- paper we will consider only the organization of general liament of India comprises two legislative bodies – the elections for electing the members of the Lok Sabha in Rajya Sabha (Upper house, corresponds to the “Sen- India. ate” in the US, or the “House of Lords” in the UK), and the Lok Sabha (Lower House, corresponds to the The total membership of the Lok Sabha is distributed “Congress” in the US, or the “House of Commons” in amongst the 35 States in such a manner that the ra- the UK). The 250 members of the Rajya Sabha are indi- tio of the population to the number of seats allotted to rectly elected by legislators of States and Union Terri- any State is nearly the same. The geographical area of tories comprising the Union of India. The 543 members the State is then demarcated into a number of territorial constituencies (with geographical boundaries), equal to Email: Bodhibrata Nag [[email protected]], Katta G. Murty the number of seats allotted, such that the population [[email protected]]. of all constituencies in that State is nearly the same.

Table 1

Lok Sabha General Elections Date of ﬁrst meeting Date of dissolution

1 25 October 1951 to 21 February 1952 13 May 1952 4 April 1957 2 24 February to 14 March 1957 10 May 1957 31 March 1962 3 19 to 25 February 1962 16 April 1962 3 March 1967 4 17 to 21 February 1967 16 March 1967 27 December 1970 5 1 to 10 March 1971 19 March 1971 18 January 1977 6 16 to 20 March 1977 25 March 1977 22 August 1979 7 3 to 6 January 1980 21 January 1980 31 December 1984 8 24 to 28 December 1984 15 January 1985 27 November 1989 9 22 to 26 November 1989 18 December 1989 13 March 1991 10 20 May to 15 June 1991 9 July 1991 10 May 1996 11 27 April to 30 May 1996 22 May 1996 4 December 1997 12 16 to 23 February 1998 23 March 1998 26 April 1999 13 5 September to 6 October 1999 20 October 1999 6 February 2004 14 20 April to 10 May 2004 2 June 2004 18 May 2009 15 16 April to 13 May 2009 1 June 2009 -

General Elections held in India

Since there are large variations in population densities government. The Chief Electoral Officer is assisted by across States, constituencies vary largely in terms of District Election Officers, Electoral Registration Offi- geographical area- thus Ladakh (the constituency with cers, and Returning Officers at the constituency level. largest area) covers 173266 sq.km in contrast to Delhi- In addition, the Election Commission co-opts a large Chandni Chowk (the constituency with smallest area) number of officials from the Central (or federal) and which covers only 11 sq.km. Each constituency has a State governments for about two months during each large number of polling stations distributed across the General Election, for conducting the elections. About 5 constituency such that voters can reach the polling sta- million officials were deployed during the 2009 General tions to cast their vote with minimum travel. The dis- Election. tribution of membership of the Lok Sabha and the to- Elections in the past have been marked by instances tal number of polling stations for each state is given in of voter intimidation through violence or harassment Table 2. in various forms, as well as clashes between political The General Elections of India are the world’s biggest opponents (Scharff [7]). These incidences have been election exercise. During the 2009 General Elections, largely arrested through deploymentof additional police a 717 million strong electorate exercised their fran- forces during the polling process in order to bring peace, chise through 1.3 million Electronic Voting Machines restore confidence in candidates and voters, and thereby deployed in 834 thousand polling stations spread across ensure fair and free elections. the length and breadth of India to elect 543 Members The Constitution of India mandates that maintenance of of the Lok Sabha from amongst 8 thousand candidates law and order is the responsibility of the States. Thus contesting the elections. The only other comparable while all States maintain police forces totaling about 1.5 elections are the European Parliament elections with an million, the average police-population ratio for all the electorate of 500 million and the US Congress elections States is only 133 police per 100,000 (National Crime with electorate of 312 million. records Bureau, 2010 [6]) in comparison with average The responsibility for conducting the elections to the international ratio of 342 (Stefan Harrendorf, 2010 [8]). Lok Sabha is vested in the Election Commission of India The Central Government therefore maintains Central according to the provisions of Article 324 of the Con- Police Forces numbering about 800 thousand under stitution of India. The Election Commission operates 7 different divisions, to complement the State police, through its secretariat based at New Delhi manned by whenever and wherever required. (Bureau of Police Re- about 300 officials. It is assisted at the State level by the search & development [1]). Chief Electoral Officer of the State, who is appointed by Since the State police are the arm of the State gov- the Election Commission in consultation with the State ernments, allegations of partisan conduct of police in Bodhibrata Nag & Katta G. Murty–Algorithmic Operations Research Vol.7 (2013) 55–70 57

Table 2

Sl.No State Number of Members Total Number of of the Lok Sabha polling stations

1 Andhra Pradesh 42 66760 2 Arunachal Pradesh 2 2057 3 Assam 14 18828 4 Bihar 40 57020 5 Goa 2 1339 6 Gujarat 26 42568 7 Haryana 10 12894 8 Himachal Pradesh 4 7253 9 Jammu & Kashmir 6 9129 10 Karnataka 28 46576 11 Kerala 20 20510 12 Madhya Pradesh 29 47812 13 Maharashtra 48 82598 14 Manipur 2 2193 15 Meghalaya 2 2117 16 Mizoram 1 1028 17 Nagaland 1 1692 18 Orissa 21 31617 19 Punjab 13 18846 20 Rajasthan 25 42699 21 Sikkim 1 493 22 Tamil Nadu 39 52158 23 Tripura 2 3008 24 Uttar Pradesh 80 129446 25 West Bengal 42 66109 26 Chattisgarh 11 20984 27 Jharkhand 14 23696 28 Uttarakhand 5 9003 29 Andaman & Nicobar Islands 1 347 30 Chandigarh 1 422 31 Dadra & Nagar Haveli 1 161 32 Daman & Diu 1 94 33 NCT of Delhi 7 11348 34 Lakshadweep 1 40 35 Puducherry 1 856

Number of constituencies and polling stations in each State enforcing law and order during the campaign closing all constituencies of those states. The days of elections phases and during the day of elections are likely. It has are spread a few days apart to allow re-deployment of therefore become universal practice to deploy Central paramilitary personnel and allow them to be familiar Police Forces (CPF), in addition to State police at all with their constituencies. For example, the 2009 Gen- polling stations during the General Elections. However, eral Election was conducted in ﬁve phases on 16 April, only about a quarter of the CPF can be spared for de- 23 April, 30 April, 7 May and 13 May. ployment during the elections, which amounts to only about 200,000 personnel. Thus General Elections are The movementof CPF personnel from their bases to the spread over different days with each day covering a few polling stations in the different phases and their sub- states only, such that the required number of CPF per- sequent return to the bases is a gigantic exercise, re- sonnel can be deployed across all polling stations of quiring coordination between different agencies such as CPF operations, Election Commission and State Chief 58 Bodhibrata Nag & Katta G. Murty–Organizing National Elections in India

Electoral Officers, District Election Officers, Railways, 2.1. The Constraints in the Problem and the Objective airlines and the Indian Air Force. In the 2009 General Function to be Optimized Election, 119 special trains, 65 sorties by Indian Air Force transport aircraft, 600 sorties by Indian Air Force While conducting the elections in phases, the Elec- helicopters and Air India chartered flights were used for tion Commission requires that the following constraints the cross-country movement of CPF personnel (Elec- should be satisfied as far as possible: tion Commission of India, 2009 [2]). 2.1.1. Constraints to be Satisfied However the process of scheduling the elections and movement of police personnel is done manually by the (a) In every State , the elections in all constituencies in Election Commission. This paper proposes an optimiza- it should be held on a single day tion methodology to find an optimum plan for organiz- (b) As far as possible, States in which elections are held ing the General Elections for all the 543 parliamentary on a day must be contiguous constituencies in the minimum possible time, with the (c) Set of States in which elections are held on consec- available CPF personnel, while minimizing the total cost utive polling days should be contiguous for the police movement involved. The paper is orga- (d) General elections over the whole country should be nized as follows: the problem statement and description completed using the smallest number of polling days of all the data is in Section 2; representation, method- (e) At every polling station, 4 CPF personnel should be ology and algorithms for the problem are described in deployed on the election day Section 3; followed by discussion of the solutions ob- (f) The total number of CPF personnel available for tained in Section 4 and conclusions in Section 5. deployment at polling stations on any polling day is at most 1,500,000 or 1.5 million. The proposed method incorporates all these constraints, in the model used to solve the problem.

2. The problem statement and description of all the 2.1.2. Selection of the objective function to optimize data In the original statement of the problem, the unit for measuring the objective function is stated in terms of The total numberof pollingstations of all the 543parlia- people-miles. But the CFP personnel deployed move mentary constituencies, spread over 35 states is 833,701. from a state to next by air (as far as possible), and from If 4 CPF (Central Police Forces) personnel are deployed the airport to the polling stations in the constituency by at each polling station, the total requirement of police road vehicles; and the costs per person-mile by air and personnel is 3.3 million if elections are to be held on road are very different. It is not logical or reasonable to the same day all over the country. However, since only add people-miles by different modes of travel directly to about a quarter of CPF amounting to 200,000 person- get the objective function to be minimized; particularly nel can be spared for deployment during the elections, since in the end all travel has to be paid for in units of so additional reserve CPF and army personnel are also money, Indian Rs (Indian Rupees or ) , in the total used to bring the number to 1.5 million for deployment travel cost of all the CFP personnel involved. during the elections. In the sequel, we will refer to this So, we use the objective function as the total cost of group of police with the responsibility of maintaining travel of all the CFP personnel involved in the General law and order at the polling booths, while elections are Elections. Air travel cost should be the cost of air travel going on there, as “CPF personne”. Even then it can be of all the moves made by air for all these personnel. seen that it is not possible to conduct elections for all Road travel can be in terms of cost incurred for it, the the 543 parliamentary constituencies on a single day. total cost of mileage of all the road vehicles used for Thus elections will have to be conducted in phases, with all these personnel. CPF personnel movement between constituencies in the phase intervals. The proposed method assumes that the 2.2. The data for the problem CPF personnel movement will be entirely by air, except the ‘last mile’ movement between the airports and the All the data for the problem can be obtained by sending constituencies. an e-mail to the ﬁrst author. Bodhibrata Nag & Katta G. Murty–Algorithmic Operations Research Vol.7 (2013) 55–70 59

2.2.1. 15 bases at Agartala, Mumbai, Kolkata, Chandigarh, The first data set required for the problem is number Delhi, Guwahati, Hyderabad, Imphal, Jaipur, Jammu, of polling stations in each constituency in each state, Jorhat, Lucknow, Patna, Shillong and Raipur. the name of the nearest airport to that constituency (CPF personnel deployed to polling stations in this constituency will use this airport to arrive in this con- 2.2.7. stituency and depart from it to their next assignment), and the road distance of this constituency to that air- The seventh data set required for the problem is the port. In the following Table 3 we show a portion of this cost of travel/person from each constituency to each information for one selected constituency in each State. constituency across the country. In the following Table 8 we show a portion of this information for a few pair 2.2.2. of selected constituencies. The second data set required for the problem is the air travel cost in (Rupees) between various pairs of airports. Actual air travel cost of CPF polling personnel is hard to get exactly since some of it is on Indian Air Force Transport Aircraft, helicopters, and some on Air 3. A graph representation of the problem Indiachartered flights. So this data is based on estimated averagecost in 10/mile(based on 2012 prices)on these A graph (also known as an undirected network) is different types of flights used, and the air distance data a pair of sets G = (N, A) where N is a set of nodes between pairs of airports. In the following Table 4 we (also called vertices, these are numbered serially and show a portion of this information for a few pairs of referenced by their numbers), and A is a set of lines airports. or edges (also called arcs in some books), each edge joining a pair of nodes. If an edge joins the pair of 2.2.3. nodes i, j it is represented by the pair (i, j) [ in some The third data set required for the problem is the es- books the symbol (i; j) is used instead]; this edge is timated cost in (Rupees) of the road or train travel of said to be incident at nodes i and j. Nodes i, j are said CPF polling personnel from airport used to the polling to be adjacent if there is an edge joining them. See stations in the constituency and back for each con- Murty[1992] for a discussion of networks and network stituency. This data is based on estimated average cost algorithms. The network G is said to be connected if for in .3/mile by these modes of travel(road or rail). In the every pair of nodes p, q in it, there is a path of edges in following Table 5 we show a portion of this information G connecting them; otherwise it is not connected. See for a few selected constituencies. Figures 6, 7. Nodes are circles with its number entered inside the circle; each edge is a straight line joining the 2.2.4. pair of nodes on it. The fourth data set required for the problem is the list of adjacent states for each state. This data is shown in full in the following Table 6. Two states are defined to be adjacent if they share a common boundary line.

2.2.5. The ﬁfth data set required for the problem is the cost in of travel/personfromthe various bases from where CPF polling personnel come from, to each constituency. In the following Table 7 we show a portion of this information for a few selected constituencies and bases.

2.2.6. Figure 6: A 6 node, 6 edge network which is not a Thesixth data set requiredforthe problemis the num- connected network (for example there is no path con- ber of personnel that come from each base. We assume sisting of edges between nodes 1, 4 in this network). 60 Bodhibrata Nag & Katta G. Murty–Organizing National Elections in India

Table 3

Sl.No State Constituency No.of Polling Nearest Distance to Stations Airport Airport (miles)

1 Andhra Pradesh Adilabad 1464 Ramagundam 85 2 Arunachal Pradesh Arunachal East 851 Pasighat 52 3 Assam Karimganj 1229 Silchar 27 4 Bihar Purvi Champaran 1193 Muzaffarpur 46 5 Goa South Goa 660 Dabolimgoa 0 6 Gujarat Anand 1510 Vadodara 22 7 Haryana Kurukshetra 1263 Chandigarh 52 8 Himachal Pradesh Mandi 1921 Kulu 20 9 Jammu & Kashmir Anantnag 1502 Srinagar 31 10 Karnataka Dharwad 1455 Hubli 12 11 Kerala Wayanad 988 Kozhikode 40 12 Madhya Pradesh Bhind 1659 Gwalior 44 13 Maharashtra Dhule 1624 Aurangabad 79 14 Manipur Inner Manipur 970 Imphal 0 15 Meghalaya Shillong 1326 Shillong 0 16 Mizoram Mizoram 1028 Aizawl 0 17 Nagaland Nagaland 1692 Dimapur 29 18 Orissa Cuttack 1319 Bhubaneswar 13 19 Punjab Jalandhar 1764 Ludhiana 34 20 Rajasthan Sikar 1574 Jaipur 63 21 Sikkim Sikkim 493 Darjeeling 30 22 Tamil Nadu Viluppuram 1376 Pondicherry 19 23 Tripura Tripura West 1558 Agartala 0 24 Uttar Pradesh Rae Bareli 1576 Lucknow 46 25 West Bengal Jalpaiguri 1560 Bagdogra 27 26 Chattisgarh Raigarh 2166 Bilaspur 81 27 Jharkhand Kodarma 2097 Gaya 43 28 Uttarakhand Garhwal 1876 Dehradun 46 29 Andaman & Nicobar Islands Andaman & Nicobar Islands 347 Portblair 0 30 Chandigarh Chandigarh 422 Chandigarh 0 31 Dadra & Nagar Haveli Dadar & Nagar Haveli 161 Daman 15 32 Daman & Diu Daman & Diu 94 Daman 0 33 NCT OF Delhi New Delhi 1540 Delhi 9 34 Lakshadweep Lakshadweep 40 Agatti 0 35 Puducherry Puducherry 856 Pondicherry 0

Number of polling stations and nearest airport of select constituencies

Table 4

AGARTALA AGATTI AGRA AHMEDABAD AIZAWL AKOLA AGARTALA - 15438 8597 11878 914 9389 AGATTI 15438 - 11907 8429 16183 7520 AGRA 8597 11907 - 4450 9485 4527 AHMEDABAD 11878 8429 4450 - 12791 3272 AIZAWL 914 16183 9485 12791 - 10276 AKOLA 9389 7520 4527 3272 10276 -

Air travel cost between select pair of airports Bodhibrata Nag & Katta G. Murty–Algorithmic Operations Research Vol.7 (2013) 55–70 61

Table 5

State/Union-Territory Parliament Number of Nearest Airport Cost of Road Travel Constituency Polling Stations from Parliament Name Constituency to nearest Airport (in Rupee/person)

Andhra Pradesh Medak 1571 HYDERABAD 289 Assam Barpeta 1396 GUWAHATI 281 Bihar Madhubani 1354 MUZAFFARPUR 274 Gujarat Surendranagar 1768 AHMEDABAD 353 Kerala Kannur 983 KOZHIKODE 389

Cost of road/train travel in Rupee/person between constituency and nearest airport, for selected constituencies.

states in India. With 1.5 million CPF personnel available for deployment for the elections, and the required 4 CPF personnel at every polling station, on a single day elections can be held at 1500000/4 = 375000 polling stations at most. Since there are 833701 polling stations over the whole country, and 833701/375000 is strictly between 2 and 3,

we need at least 3 polling days to complete (1) Figure7:A 6 node,7 edgenetworkthat is aconnected network holding the elections over the whole country. A partial network of N is a network G1 = (N1, A1) where the set of nodes N1 is a subset of N, and the set So, on any single day, elections are held in a subset of of edges N1 is the set of all edges of G that have both states in the country. Suppose elections are held on k their incident nodes in N1.G1 is said to be the partial different days. We know that k >= 3. Let NIt be the set network of G induced by the subset of nodes N1, it is of states in which elections are held on the tth day for connected if it forms a connected network. For example t = 1 to k, and let GIt = (NIt, AIt) be the partial net- for the network G in Figure 5, G1 = ({1, 2, 3}, {(1, work of GI induced by the subset of nodes NIt. So, our 2), (2, 3), (3, 1)}) is a connected partial network; and problem boils down to ﬁnding the partition (GI1, ..., G2 = ({1, 2, 3, 4}, {(1, 2), (2, 3), (3, 1)}) is a partial GIk) of GI satisfying constraints (a) to (f) listed above network which is not connected because there is no path in Section 2.1.1, and sequence these partial networks connecting nodes 1 and 4 in it. generated corresponding to the election days 1, 2, ...,

Let {N1, N2, ..., Nk} be a partition node set N in G k; while minimizing the total cost of moving the CPF (i.e., their union is G, and every pair of subsets in this personnel across the country to monitor the polling sta- partition are mutually disjoint). For t =1to k, let Gt = tions as required. So our problem actually involves a (Nt,At) be the partial network of G induced by Nt. Then graph partitioning problem, and a sequencing problem. {G1,...,Gk} is said to be a partition of G. The graph In constraint (b) of Section 2.1.1, the meaning of the partitioning problem is the general problem of finding word “contiguous” is left somewhat vague. We will in- a partition of a given graph G subject to specified con- terpret it to mean that the partial network GIt of GI in- straints that minimizes a specified objective function. duced by the set of states in which elections are held on Our polling problem can be viewed as an instance of the tth day should be a connected network for each t = the graph partitioning problem. To see this let NI = {1, 1 to k. Also, we will interpret constraint (c) of Section ..., 35} where node i represents the state of India with 2.1.1 that the set of states in which elections are held on serial number i in Table 2. Let AI = {(i, j): nodes i, j consecutive days should be “contiguous” to mean that correspond to adjacent states as given in Table 4}. Then there should be at least one edge in GI joining a node (NI, AI) = GI is known as the adjacency network for in NIt to a node in NIt+1 for each t =1 to k-1. 62 Bodhibrata Nag & Katta G. Murty–Organizing National Elections in India

Table 6

State Adjacent States

Andhra Pradesh Orissa,Chattisgarh, Maharashtra, Karnataka, Tamil Nadu Arunachal Pradesh Assam Assam West Bengal, Arunachal Pradesh, Nagaland, Manipur, Tripura, Meghalaya, Mizoram Bihar Uttar Pradesh, Jharkhand, West Bengal Goa Karnataka, Maharashtra Gujarat Maharashtra, Madhya Pradesh, Dadra & Nagar Haveli, Daman & Diu, Rajasthan Haryana Punjab, NCT-Delhi, Chandigarh, Himachal Pradesh, Uttar Pradesh, Rajasthan Himachal Pradesh Punjab, Jammu & Kashmir, Uttar Khand, Haryana, Chandigarh Jammu & Kashmir Punjab, Himachal Pradesh Karnataka Goa, Maharashtra, Andhra Pradesh, Tamil Nadu, Kerala, Lakshadweep Kerala Karnataka, Tamil Nadu, Lakshadweep Madhya Pradesh Maharashtra, Gujarat, Rajasthan, Uttar Pradesh, Chattisgarh Maharashtra Goa, Karnataka, Andhra Pradesh, Chattisgarh, Madhya Pradesh, Gujarat, Daman & Diu, Dadra & Nagar Haveli Manipur Nagaland, Mizoram, Assam Meghalaya Assam Mizoram Assam, Tripura, Manipur Nagaland Manipur, Assam Orissa Andhra Pradesh, Chattisgarh, Jharkhand, West Bengal Punjab Rajasthan, Haryana, Himachal Pradesh, Jammu & Kashmir, Chandigarh Rajasthan Gujarat, Madhya Pradesh, Uttar Pradesh, Haryana, Punjab Sikkim West Bengal Tamil Nadu Kerala, Karnataka, Andhra Pradesh, Puducherry Tripura Assam, Mizoram Uttar Pradesh Uttarakhand, Himachal Pradesh, Haryana, Rajasthan, NCT-Delhi, Madhya Pradesh, Chattisgarh, Jharkhand, Bihar West Bengal Orissa, Jharkhand, Bihar, Sikkim, Assam Chattisgarh Orissa, Andhra Pradesh, Maharashtra, Madhya Pradesh, Uttar Pradesh, Jharkhand Jharkhand Orissa, Chattisgarh, Uttar Pradesh, Bihar, West Bengal Uttarakhand Himachal Pradesh, Uttar Pradesh Andaman & Nicobar Islands Tamil Nadu, Andhra Pradesh Chandigarh Punjab, Haryana, Himachal Pradesh Dadra & Nagar Haveli Gujarat, Maharashtra Daman & Diu Gujarat, Maharashtra NCT of Delhi Haryana, Uttar Pradesh Lakshadweep Kerala, Goa, Karnataka Puducherry Tamil Nadu

Adjacent states of each state

Table 7

State/Union-Territory ParliamentConstituency Central Police Force Bases Agartala Mumbai Chandigarh Kolkata Andhra Pradesh Peddapalle 8474 4353 6425 8505 Jammu & Kashmir Baramulla 12377 10652 11615 2756 Karnataka Bijapur 11164 2579 9120 9409 Madhya Pradesh Tikamgarh 7605 6471 5983 4917 Punjab Amritsar 11385 8812 10390 1290

Cost of travel in Rs./person from a few selected bases to some constituencies Bodhibrata Nag & Katta G. Murty–Algorithmic Operations Research Vol.7 (2013) 55–70 63

Table 8

Hassan Vidisha Mayurbhanj Sangrur Tura (Karnataka) (Madhya Pradesh) (Orissa) (Punjab) (Meghalaya)

Medak 4390 4595 6751 10048 10598 (Andhra Pradesh) Nagpur 6703 2000 4920 7305 7953 (Maharashtra) Uluberia 10966 7286 1704 10000 3046 (West Bengal) Fatehpur 10672 3307 4885 4676 6421 (Uttar Pradesh) Nowgong 14938 10235 5270 11064 2231 (Assam)

Cost of travel in Rs./person between selected pairs of constituencies

3.1. A Hamiltonian Path Heuristic For the Problem minimum cost Hamiltonian path in GI, and then divide it into segments of required size, and take the partial A Hamiltonian path in an undirected network G is a networks of GI to be the partial networks induced by path consisting of edges in the network that contains the sets of nodes of the various segments. This is the all the nodes in the network. Given the lengths of all basis for this method, which we will describe next. The the edges in G, the Hamiltonian path problem in G method has two stages. Stage 1 determines the election is that of finding a shortest Hamiltonian path in G. A day for each state with day 1 as the starting day of the Hamiltonian path in G is a path in G that contains all elections. the nodes in G; and a shortest Hamiltonian path is one Selecting the sequenceof partial networksof GI to cover whose length (the length of a Hamiltonian path is the the various polling days as the sequence of segments sum of the lengths of the edges in it) is the shortest along the shortest Hamiltonian path helps keep the cost among all Hamiltonian paths. In some applications, we of movements of the NPF polling team between con- may require Hamiltonian paths beginning with a spec- secutive polling days small, thus achieving our objec- ified node as the initial node. This can easily be trans- tive of minimizing the total cost of travel of this team in formed into the well known traveling salesman prob- conducting the elections, while satisfying the contiguity lem. requirements (b), (c) among the specified constraints of Since our CPF polling personnel team has to cover each Section 2.1.1. of the states in India satisfying the constraints listed above, the shortest Hamiltonian path in the network GI 3.1.1. Stage 1: Determining the Election Day for Each using the cost of traveling/person between states i, j as State the length of edge (i, j) , may provide useful information There may be up to 13 different airports that the CPF to develop a good solution to our problem. But there are polling team will use to enter and leave a state. In reality important differences. No member of the CPF polling then, each individualof this team has a separate problem team visits all the states, everyone visits only a subset to be solved to estimate correctly the cost incurred for of the states, so our problem is really one of partitioning his participation; but all the data for that individual is GI into connected partial networks GI1,...,GIk of GI, not known until the solution of this stage is determined, corresponding to election days 1 to k such that there is making it impractical to get into such precise detail. an edge joining a node in GIt to a node in GIt+1 for So, as an approximation, we will assume that all mem- all t = 1 to k-1; and each member of the CPF polling bers of the CPF team visiting a state will travel to and team visits one node in each of GIt for t =1to k. from this state use one airport, the one that majority of Define a segment of a Hamiltonian path as the por- this team will use (this is a reasonable and good ap- tion of this path between a pair of nodes on it. One proximation). way of generating partial networks of GI satisfying the So, contiguity requirements in our problem is to obtain a 64 Bodhibrata Nag & Katta G. Murty–Organizing National Elections in India make the cost coefficient of edge (i, j) in GI to be Re-numbering of the states: Re-number the states in cij = cost of air travel/person between the airports the order of appearance on H beginning with number 1 (determined as above) corresponding to states i, j. for the state “NCT of Delhi”, the starting state. In the

For each node i in GI define its weight wi = number sequel, we will refer to the states by these numbers. of polling stations in state i. As the polling team moves from one state to the other, 3.1.1.2 Step 2: Determining the Election Day for our original problem is actually many Hamiltonian path each state: problems, one for each individual in the CPF polling team, depending on the airports they will use for each Define move. By using a single airport for each state, one that majority will use, we will adopt a single countrywide Weight of a segment of H = sum of the weights shortest Hamiltonian path as an approximation to the wi of nodes on it collection of all of them. This Stage 1 involves two Steps, we will describe them now. wo = maximum number of polling stations (2) that the CPF polling team can handle in a day 3.1.1.1 Step 1: Finding the Shortest Hamiltonian = 1500000/4 = 375000 Path in GI In this step we will break up H into the smallest number We assume that the national elections in India always of segments subject to the constraint that the weight begin in the Nation’s Capital, “NCT of Delhi” (State of each segment is <= w0; each segment corresponds number 33). So, find a shortest Hamiltonian path in to an election day in the solution developed by this GI beginning with State 33 as the starting state, and method. We discuss two algorithms for breaking up covering all the airports selected above corresponding H into segments. This problem is known as the tour to the various states. Let this Hamiltonian path be H. segmentation problem. The order of states on H, is the order in which elections will be held in the solution determined by this method. Algorithm 1: A Simple Heuristic Algorithm for Breaking Up H Into Segments In this algorithm segments are formed one by one. Start- ing at node1, movealong H until the sum of the weights of states included so far is <= w0, and exceeds w0 if the next state is included. At this time complete the current segment, remove it from H, and continue the same way with the remaining part of H. Let k be the number of segments obtained by this heuristic. Number these segments 1 to k in the order of appearance on H. The solution obtained by this heuristic is to hold elections in all the states in segment p on day p, for each p =1to k. We applied this algorithm and found the solution to the problem, k in it is = 3. From (1), we conclude that this solution is optimal to the problem of breaking up H into the smallest number of segments subject to the constraint that the weight of each segment should be <= w0 . According to this solution, the elections will be held according to the following schedule:

Figure 8: Shortest Hamiltonian Path covering the • Day 1: NCT-Delhi, Haryana, Punjab, Jammu & airports used by the CPF team for traveling to and from Kashmir, Himachal Pradesh, Chandigarh, Uttarak- the states. hand, Uttar Pradesh, Bihar, Sikkim, Jharkhand, West Bengal, Meghalaya, Assam, Arunachal Pradesh, Bodhibrata Nag & Katta G. Murty–Algorithmic Operations Research Vol.7 (2013) 55–70 65

Table 9

Arc Start node on arc i End node on arc i Length of no. i arc i on on H H (miles) State/Union Territory no. Airport State/Union Territory no. Airport i = 1 1 =NCT Delhi Delhi 2 = Haryana Delhi 0

2 2 = Haryana Delhi 3 = Punjab Ludhiana 178 3 3 = Punjab Ludhiana 4 = Jammu & Kashmir Srinagar 228 4 4 = Jammu & Kashmir Srinagar 5 = Himachal Pradesh Kulu 199 5 5 = Himachal Pradesh Kulu 6 = Chandigarh Chandigarh 86 6 6 = Chandigarh Chandigarh 7 = Uttarkhand Dehradun 70 7 7 = Uttarkhand Dehradun 8 = Uttar Pradesh Lucknow 298 8 8 = Uttar Pradesh Lucknow 9 = Bihar Muzaffarpur 279 9 9 = Bihar Muzaffarpur 10 = Sikkim Darjeeling 189 10 10 = Sikkim Darjeeling 11 = Jharkhand Ranchi 316 11 11 = Jharkhand Ranchi 12 = West Bengal Kolkata 201 12 12 = West Bengal Kolkata 13 = Meghalaya Shillong 305 13 13 = Meghalaya Shillong 14 = Assam Guwahati 46 14 14 = Assam Guwahati 15 = Arunachal Pradesh Zero 160 15 15 = Arunachal Pradesh Zero 16 = Nagaland Dimapur 116 16 16 = Nagaland Dimapur 17 = Manipur Imphal 77 17 17 = Manipur Imphal 18 = Mizoram Aizawl 108 18 18 = Mizoram Aizawl 19 = Tripura Agartala 91 19 19 = Tripura Agartala 20 = Andaman & Nicobar Islands Port Blair 847 20 20 = Andaman & Nicobar Islands Port Blair 21 = Orissa Bhubaneswar 751 21 21 = Orissa Bhubaneswar 22 = Chattisgarh Raipur 281 22 22 = Chattisgarh Raipur 23 = Andhra Pradesh Hyderabad 337 23 23 = Andhra Pradesh Hyderabad 24 = Tamil Nadu Chennai 322 24 24 = Tamil Nadu Chennai 25 = Puducherry Pondicherry 84 25 25 = Puducherry Pondicherry 26 = Karnataka Bangalore 165 26 26 = Karnataka Bangalore 27 = Kerala Kochi 229 27 27 = Kerala Kochi 28 = Lakshwadeep Agatti 285 28 28 = Lakshwadeep Agatti 29 = Goa Dabolim Goa 332 29 29 = Goa Dabolim Goa 30 = Maharashtra Mumbai 264 30 30 = Maharashtra Mumbai 31 = Dadra & Nagar Haveli Daman 93 31 31 = Dadra & Nagar Haveli Daman 32 = Daman & Diu Daman 0 32 32 = Daman & Diu Daman 33 = Gujarat Ahmedabad 182 33 33 = Gujarat Ahmedabad 34 = Madhya Pradesh Indore 211 34 34 = Madhya Pradesh Indore 35 = Rajasthan Jaipur 291

Arcs in the Shortest Hamiltonian Path H in Figure 7 in order of appearance on H

Nagaland, Manipur, Mizoram- total 18 States, 245 per bound on the minimum number of segments into constituencies, 373,574 polling stations which H can be partitioned subject to the upper bound • Day 2: Tripura, Andaman & Nicobar Islands, w0 on the weight of each segment. From the results Orissa, Chattisgarh, Andhra Pradesh, Tamil Nadu, obtained from the solution given by Algorithm 1, we Puducherry, Karnataka, Kerala, Lakshwadeep, Goa, know that k = 3 is an upper bound that we need. We Maharashtra, Dadra & Nagar Haveli, Daman & Diu, will describe this model denoting this upper bound by Gujarat- total 15 States, 244 constituencies,369,616 k (in our problem k = 3). Deﬁne variables polling stations • Day 3: Madhya Pradesh, Rajasthan- total 2 States, 54 yi,j = 1 if State i belongs to segment j , for i = 1 to constituencies, 90,511 polling stations 35, j = 1 to k; Algorithm 2: A 0-1 model for breaking up H into 0, otherwise. segments: For constructing this model we need an up- 66 Bodhibrata Nag & Katta G. Murty–Organizing National Elections in India

For each j, we need to form constraints to guarantee k yi,j ≤ 1 for each i =1 to 35, that the set of all states i corresponding to yi,j = 1 form X a segment. For this, for each j and each i = 1to33we j=1 must guarantee that: 35 X wiyi,j ≤ w0 for each j =1 to k, i=1 yi,j = yi+2,j = 1 implies that yi+1,j = 1 also. yi,j + yi+2,j − yi+1,j ≤ 1 for all i =1 to 33, and j =1 to k, This requires that if yi,j +yi+2,j = 2 then yi+1,j = 1; y1,1 =1, k k k and if yi,j +yi+2,j = 1 then yi+1,j =0 or 1. X jyi−1,j ≤ X jyi,j ≤ 1+ X jyi+1,j j=1 j=1 j=1 So, for the pair (yi,j + yi+2,j ; yi+1,j ) all the values in for all i =2 to 35. the set {(0, 0), (0, 1), (1, 0), (1, 1), (2, 1)} are possible, all y =0 or 1 for all i =1 to 35, and j =1 to k. but (2, 0) is not. i,j Here is an explanation for these constraints. From the Plotting these points on the 2-dimensional Cartesian definition of the binary variables yij given above, the plane with yi,j + yi+2,j on the horizontal axis and yi+1,j first constraint here guarantees that each of the states on the vertical axis; we will now see see that yi,j + i = 1 to 35 is contained in exactly one segment. The yi+2,j , and yi+1,j must satisfy the constraint yi,j + second constraint guarantees that the weight of each yi+2,j - yi+1,j = 1. To see this clearly, denote: segment obtained is = the maximum possible weight w0 determined above in (2). It is discussed above that the third constraint guarantees that for each j the set of all the states i corresponding to yij = 1 form a segment.The fourth constraint says that Segment 1 begins with State 1 where the elections begin in the country on the first day of polling. The fifth constraint says that for each i = 2 to 35, either State i belongs to the same segment as State i - 1, or the next segment following it.The last constraint says that all the variables are binary variables. Clearly these are all the constraints in the problem of breaking up H into segments. From the numbering of the states (defined above in Ta- ble 9), this guarantees that each segment is a portion Figure 9: Plot of feasible values for the integer vari- of the Hamiltonian path H between two nodes, and that ables x (= y + y ) and x (= y ). All feasible 1 i,j i+2,j 2 i+1,j segments are numbered serially in the order in which values {(x1, x2):(x1, x2) = (0,0), (0,1), (1, 0), (1,1), they appear along H. (2,1)} are plotted with a “•”, and all of them satisfy x1 Now the actual number of segments into which H is -x2 <= 1. partitioned in the solution obtained, is the serial number of the segment containing the last State 35 on H; k so this is jy35 , which we have to minimize. So y +y by x , and Pj=1 ,j i,j i+2,j 1 the binary integer programming model for this tour seg- yi+1,j by x2 k mentation problem is to minimize Pj=1 jy35,j subject to the above constraints. and by plotting feasible values of these variables on the We solved this 0-1 integer programming problem and x1,x2-Cartesian plane we get the following Figure 9. obtained an optimum solution, consisting of 3 segments. According to it, the elections will be held by the following schedule: From this we see that the system of constraints that the variables yi,j have to satisfy are: • Day 1: NCT-Delhi, Haryana, Punjab, Jammu & Bodhibrata Nag & Katta G. Murty–Algorithmic Operations Research Vol.7 (2013) 55–70 67

Kashmir, Himachal Pradesh, Chandigarh, Uttarak- batches of polling personnel required to hold elections hand, Uttar Pradesh, Bihar, Sikkim, Jharkhand, West at the corresponding state on day j+1 . An optimum Bengal, Meghalaya, - total 13 States, 225 constituen- solution of this bipartite minimum cost ﬂow problem cies, 347,776 polling stations (a transportation problem) gives the moves to be made • Day 2: Assam, Arunachal Pradesh, Nagaland, Ma- for this transitions of the polling personnel from day j nipur, Mizoram, Tripura, Andaman & Nicobar Is- to day j+1 for each j =1 to k -1. lands, Orissa, Chattisgarh, Andhra Pradesh, Tamil Nadu - total 11 States, 136 constituencies, 200,672 Moving CPF polling personnel from their bases to polling stations states in segment 1 on Day 1, and from states in the • Day 3: Puducherry,Karnataka, Kerala, Lakshwadeep, last segment back to their bases at the end of elec- Goa, Maharashtra, Dadra & Nagar Haveli, Daman & tions: This problem can also be modeled as a bipar- Diu, Gujarat, Madhya Pradesh, Rajasthan- total 11 tite minimum cost ﬂow problem exactly as above, and States, 182 constituencies, 285,253 polling stations. solved.

Like the Heuristic Algorithm 1, the solution obtained by Algorithm 2 also partitions the shortest Hamiltonian 4. Results path H into 3 segments, which is the minimum possible. Now we will take the best among the solutions ob- Solving the model, we obtain a solution with a cost of tained from both the methods as the one to implement. about 25 billion (Rupees). The model takes about 21 Even though both correspond to the same value 3 for seconds for processing and solution using IBM ILOG the number of segments into which H is partitioned; CPLEX 12.1.0 on a 1.6 GHz computer. the one obtained by Algorithm2 is better than that ob- The optimal movement of Central Police Force person- tained by Algorithm1, in other respects. For example, nel from Agartala base is given in Table 10 as an il- in the solution obtained by Algorithm 2, the number of lustration. Here “number moved” is the number of po- constituencies and polling stations are more evenly dis- lice moving from a base/constituency to another con- tributed across the segments. Hence we will implement stituency in the next segment. the solution obtained by Algorithm2. “Number required” is the number of police personnel In the sequel k denotes the numberof segments selected, required at that constituency (given by 4 times the num- which is 3. ber of polling stations in that constituency). Hence there are 2 situations occurring: 3.1.2. Stage 2: Moving polling personnel from one - number moved exactly equals the number required state to the next during the elections which means that this set of movements is sufﬁcient to meet its needs.For example,in base to segment1,move- After completing the elections in the states in segment ment from Agartala Base to West Bengal-Dum Dum j on day j, the batches of polling personnel have to be (5992/5992which means that 5992personnelare moved moved to the states in segment j+1 for j =1to k-1. and positioned at that constituency while 5992 person- Construct a bipartite network F with states in segment nel are required at that constituency as per norms). j as the source nodes, and states in segment j+1 as sink - number moved is less than number required, which nodes. Join each source node p to each sink node q 2 means that there are policemen coming in from other by a directed arc (p, q) with cost coefﬁcient c p,q = places to meet the requirement in this constituency. plane-fare between states p, q using the appropriate For example, in segment 2 to segment 3 movements, airports in each state to minimize the cost for this move Karnataka- Chikkballapur requires 7292 police which (at this stage with information available we could use is met by 1788 from Tamil Nadu-Arani and 5504 from different airports than those used in Stage 1 to make Tamil Nadu-Viluppuram. the total cost for this stage small. Also at this stage, notice that the airports used by different batches of polling personnel may be different in each state. Make 5. Discussion availability (quantity to be shipped to the set of sink nodes) at each source node = the number of batches of The method demonstrated in Section 4, enables (a) polling personnel in the corresponding state on day j, scheduling of elections within the minimum number and the requirement at each sink node = the number of of segments (b) sequencing the segments, such that the 68 Bodhibrata Nag & Katta G. Murty–Organizing National Elections in India

Table 10

To West Bengal State 1.Barrackpore (3164/5308) Constituencies: Number 2.Dum Dum(5992/5992) moved/ Total Number 3.Barasat(6128/6128) required 4.Joynagar(5700/5700) 5.Jadavpur(6516/6516) 6.Kolkata Dakshin(7452/7452) 7.Kolkata Uttar(6584/6584) 8.Howrah(6600/6600) 9.Jhargram(7016/7016) 10.Purulia(6336/6336) 11.Bankura(6752/6752) 12.Bishnupur(6488/6488) 13.Burdwan Durgapur(6752/6752) 14.Asansol(6252/6252)

To Jharkhand State 1.Dumka(396/6572) Constituencies: Number 2.Jamshedpur(6504/6504) moved/ Total Number 3.Singhbhum(5368/5368) required Movement of Agartala based Central Police Force personnel from base to Segment 1 movement of Central Police Forces (measured in men- References miles) is minimized and (c) sourcing the appropriate number personnel from the most convenient bases. [1] Bureau of Police Research & Development. (n.d.). Data The method assumes that there is only one set of move- on Police Organizations Retrieved November 6, 2011, from BPRD(Ministry of Home Affairs, Government ments from the bases, which is from the bases to con- of India): http://www.bprd.nic.in/index3. stituencies where the first segments of elections are be- asp?sslid=547\&\\subsublinkid=204\ ing held. Similarly there is only one set of movements &lang=1, 2009. from the segment where the final phases of elections are [2] Election Commission of India. (2009). Lok Sabha being held to the bases. There are no movements be- Election 2009-Reinforcing Indian Democracy. New tween the bases and any segment, where intermediate Delhi: Election Commission of India. phases of elections are in progress. [3] Google. (n.d.). Google Maps. Retrieved October 2, 2011, http://maps.google.co.in The method can be modified to incorporate ground re- from . [4] iTouchMap.com. (n.d.). Latitude and Longitude of a alities. For example, the requirement of polling person- point. Retrieved October 2, 2011, from http:// nel may vary across constituencies, depending on the itouchmap.com/latlong.html. perceptions of threat to maintenance of law and order. [5] Murty, Katta G., (1992), Network Pringogramming, In that case, suitable data will have to be incorporated Prentice-Hall, Englewood Cliffs, NJ, http://ioe. in both the models. engin.umich.edu/people/fac/books/ murty/network\_programming/ also can be seen at this website. [6] National Crime Records Bureau. (2010). Crime in India 6. Conclusion 2009 Statistics. New Delhi: NCRB (Ministry of Home Affairs, Government of India). In this paper, a methodology is proposed and demon- [7] Scharff, M. (n.d.). Policing Election Day: Vulnerability strated for obtaining the optimal scheduling and lo- Mapping in India 2006-2009. Retrieved November gistics planning of the Indian General Elections. The 6, 2011, from http://www.princeton.edu/ successfulsocieties/content/data/ method can be utilized for scheduling and planning any policy\_note/PN\_id173/Policy\_Note\ nation-wide event requiring scarce resources. _ID173.txt. [8] Stefan Harrendorf, M. H. (2010). International Statistics Here “n.d.” means “no date”. on Crime and Justice. Helsinki: European Institute for Bodhibrata Nag & Katta G. Murty–Algorithmic Operations Research Vol.7 (2013) 55–70 69

Table 11

Movement from West Bengal State Constituencies to 1.Barrackpore to Chennai North(1288/4968) Tamil Nadu State Constituencies (Number moved/ Total 2.Barrackpore to Chennai South(1876/5576) Number required) 3.Dum Dum to Mayiladuthurai(5292/5292) 4.Barasat to Chennai North(1536/4968) 5.Barasat to Chennai Central(4592/4592) 6.Jadavpur to Chidambaram(2944/5612) 7.Kolkata Dakshin to Arani(5548/5548) 8.Kolkata Uttar to Arakkonam(5388/5388) 9.Howrah to Viluppuram (5504/5504) 10.Howrah to Chidambaram(1096/5612)

Movement from West Bengal State Constituencies to 1.Dum Dum to Bhadrak(700/6168) Orissa State Constituencies (Number moved/ Total Num- 2.Joynagar to Kandhamal(56/5152) ber required) 3.Joynagar to Kendrapara(336/6656) 4.Joynagar to Jagatsinghpur(5308/6844) 5.Jadavpur to Dhenkanal(3572/5500) 6.Kolkata Dakshin to Aska(1904/5560) 7.Kolkata Uttar to Jajpur (1196/5596) 8.Jhargram to Balasore(5856/5856) 9.Purulia to Sambalpur (5604/5604) 10.Bankura to Nabarangpur(5828/5828) 11.Bishnupur to Mayurbhanj(6488/6488) 12.Burdwan Durgapur to Keonjhar(332/6584) 13.Burdwan Durgapur to Koraput(1612/6064) 14.Asansol to Keonjhar(6252/6584)

Movement from West Bengal State Constituencies to 1.Jhargram to Srikakulam(1160/7160) Andhra Pradesh State Constituencies (Number moved/ 2.Purulia to Aruku (732/6424) Total Number required) 3.Bankura to Aruku (924/6424) 4.Burdwan Durgapur to Kakinada(4808/5780)

Movement from Jharkhand State Constituencies to 1.Dumka to Aruku(396/6424) Andhra Pradesh State Constituencies (Number moved/ 2.Jamshedpur to Srikakulam(692/7160) Total Number required) 3.Jamshedpur to Kakinada(420/5780) 4.Jamshedpur to Narsapuram(5392/5392) 5.Singhbhum to Anakapalli(5368/6220) Movement of Agartala based Central Police Force personnel from Segment 1 to Segment 2

Crime Prevention & Control (HEUNI). [9] The Airport Authority. (n.d.). Every airport, everywhere. Retrieved November 8, 2011, from http:// Received 4-1-2013; accepted 18-2-2013 airport-authority.com/ 70 Bodhibrata Nag & Katta G. Murty–Organizing National Elections in India

Table 12

Movement from Andhra Pradesh State Constituencies 1.Aruku to Akola(2052/6740) to Maharashtra State Constituencies (Number moved/ 2.Srikakulam to Buldhana(1852/6680) Total Number required) 3.Kakinada to Palghar(420/7696) 4.Narsapuram to Bhiwandi(5392/7304) 5.Kakinada to Palghar(3868/7696) 6.Kakinada to Bhiwandi(940/7304)

Movement from Andhra Pradesh State Constituencies 1.Anakapalli to Bhavnagar(4272/6176) to Gujarat State Constituencies (Number moved/ Total 2.Anakapalli to Rajkot(1096/6324) Number required)

Movement from Orissa State Constituencies to Ma- 1.Nabarangpur to Yavatmal Washim(5828/7524) harashtra State Constituencies (Number moved/ Total 2.Koraput to Amravati(1612/7180) Number required) 3.Bhadrak to Ramtek(700/8180)

Movement from Orissa State Constituencies to Mad- 1.Balasore to Rewa(5856/5808) hya Pradesh State Constituencies (Number moved/ Total 2.Sambalpur to Rewa (5604/5808) Number required) 3.Mayurbhanj to Shahdol(6488/6664) 4.Keonjhar to Rewa (6584/5808) 5.Kandhamal to Rewa (56/5808) 6.Kendrapara to Rewa (336/5808) 7.Jagatsinghpur to Rewa (5308/5808) 8.Dhenkanal to Rewa (3572/5808) 9.Aska to Rewa (1904/5808) 10.Jajpur to Balaghat(1196/7016)

Movement from Tamil Nadu State Constituencies to 1.Chennai North to Bangalore Central(1744/7068) Karnataka State Constituencies (Number moved/ Total 2.Chennai North to Bangalore South (956/6996) Number required) 3.Chennai North to Kolar(124/7560) 4.Chennai South to Bangalore North(1876/7804) 5.Mayiladuthurai to Hassan(4368/6000) 6.Mayiladuthurai to Bangalore North(736/7804) 7.Mayiladuthurai to Shimoga(188/6820) 8.Chennai Central to Davangere(3204/6620) 9.Chennai Central to Tumkur(1232/6844) 10.Chennai Central to Bangalore Rural(156/8644) 11.Chidambaram to Tumkur (4040/6844) 12.Arani to Bangalore Rural(336/8644) 13.Arani to Chikkballapur(1788/7292) 14.Arakkonam to Bagalkot(32/6012) 15.Viluppuram to Chikkballapur (5504/7292)

Movement from Tamil Nadu State Constituencies to 1.Arani to Puducherry(3424/3424) Puducherry State Constituencies (Number moved/ Total Number required) Movement from Tamil Nadu State Constituencies to Goa 1.Arakkonam to North Goa(2716/2716) State Constituencies (Number moved/ Total Number re- 2.Arakkonam to South Goa(2640/2640) quired) Movement of Agartala based Central Police Force personnel from Segment 2 to Segment 3